$\int_{23\pi}^{71\pi/2}\ln \left ( \frac{\left ( 1+\sin x \right )^{1+\cos x}}{1+\cos x} \right )\,dx$ I ran into this integral and I'm trying to solve it.
$$I=\int_{23\pi}^{71\pi/2}\ln \left ( \frac{\left ( 1+\sin x \right )^{1+\cos x}}{1+\cos x} \right )\,dx$$
Well, this has something to do with symmetries.
At first I applied the sub $u=23\pi+71\dfrac{\pi}{2}-x$ but I didn't like the result that this sub gave me, so I didn't dwelve further. Then I thought of applying log properties so that the integral was rewritten as:
$$\begin{aligned}
&\int_{23\pi}^{71\pi/2}\ln \left ( \frac{\left ( 1+\sin x \right )^{1+\cos x}}{1+\cos x} \right )\,dx\\ &=\int_{23\pi}^{71\pi/2}\left[ \left ( 1+\cos x \right )\ln \left ( 1+\sin x \right )- \ln \left ( 1+\cos x \right )\right ] \,dx\\ 
 &= \int_{23\pi}^{71\pi/2}\left ( 1+\cos x \right )\ln \left ( 1+\sin x \right )dx-\int_{23\pi}^{71\pi/2}\ln \left ( 1+\cos x \right )\,dx
\end{aligned}$$
then substite $u=1+\sin  x$ or $u=1+\cos x$ change the limits of integration, but then again I don't get much help out of it, because the limits are bizzare (at least to me). If the limits were OK, then I would apply IBP to both integrals (definitely not a good way to go around), however I have not also dwelved in it further. I also didn't check if by splitting the integral apart I get into improper ones, or one of them is not defined on the given interval.
Any ideas?
Edit: Some typo correction was done!
 A: Note that
$$
\int\left[\vphantom{\sum}\log(1+\sin(x))-\log(1+\cos(x))\right]\mathrm{d}x\tag{1}
$$
is $0$ over quadrants $1$ and $3$ and is opposite over quadrants $2$ and $4$. Thus, the integral is $0$ over the whole circle.
Furthermore,
$$
\begin{align}
\int\cos(x)\log(1+\sin(x))\,\mathrm{d}x
&=\int\log(1+\sin(x))\,\mathrm{d}(1+\sin(x))\\
&=(1+\sin(x))\log(1+\sin(x))-\sin(x)+C\tag{2}
\end{align}
$$
Thus, this integral is $0$ over the whole circle as well.
Note that your integral is the sum of $(1)$ and $(2)$
The domain of integration is $[23\pi,35\pi]\cup[35\pi,71\pi/2]$. $[23\pi,35\pi]$ consists of $6$ full circles and $[35\pi,71\pi/2]$ is quadrant $3$. Thus, we only need compute $(2)$ over quadrant $3$, and that is $1$.
Therefore,
$$
\int_{23\pi}^{71\pi/2}\log\left(\frac{(1+\sin(x))^{1+\cos(x)}}{1+\cos(x)}\right)\,\mathrm{d}x=1\tag{3}
$$
A: Some hints:


*

*Since $\sin$ and $\cos$ are periodic with period $T=2\pi$, $\int_T^{a+T} f(\sin x,\cos x)dx=\int_0^af(\sin x,\cos x)dx$

*$\int_a^b f(x)dx=\int_a^b f(a+b-x)dx$ which is particulary useful with trigonometric functions.

*$\int_0^2a f(x)dx=\int_0^a f(x)dx+f(2a-x)dx$



Part of your Integral:
$$I=\int_{23\pi}^{71\pi/2}\ln(1+\cos x)dx\\I=\int_0^{25\pi/2}\ln(1+\cos x)dx\\I=12\int_0^{\pi}\ln(1+\cos x)dx+\int_0^{\pi/2}\ln(1+\cos x)dx\\I=12\int_0^{\pi}\ln(1-\cos x)dx+\int_0^{\pi/2}\ln(1+\sin x)dx\\2I=12\int_0^{\pi}\ln((1+\cos x)(1-\cos x))+2\int_0^{\pi/2}\ln(1+\cos x)dx\\
2I=24\int_0^{\pi}\ln(\sin x)+2\int_0^{\pi/2}\ln(1+\cos x)dx$$

Intermediate:
$$\int_0^{\pi}\ln(\sin x)\\=\int_0^{\pi/2}\ln(\sin x)+\ln(\sin (\pi/2-x))\\=\int_0^{\pi/2}\ln(\sin 2x)-\ln(2)\\=0.5\int_0^{\pi}\sin(x)dx-\pi\ln(2)/2$$

So, $$I=12(-\pi\ln(2))+2\int_0^{\pi/2}\ln(1+\cos x)$$
Now your integral is(let it be J):
$$J=I+\int_0^{25\pi/2}(1+\cos x)\ln(1+\sin x)\\
\small K(\text{let})=J+\pi\ln(2)=2\int_0^{\pi/2}\ln(1+\cos x)+12\int_0^{\pi}(1+\cos x)\ln(1+\sin x)+\int_0^{\pi/2}(1+\cos x)\ln(1+\sin x)\\=\ldots\\$$
Hereafter make it in form $\int(P(x))\ln(1+\cos x)$. I hope you can do from here?
